Simple derivative. Where is the logical fallacy here?

[PinkCrayon]

$2^{2} = 2 + 2$
$3^{2} = 3 + 3 + 3$
$4^{2} = 4 + 4 + 4 + 4$

Generalizing...
$x^{2} = x + ... + x$ (x times)

Take the derivative of both sides.
$\frac{\delta}{\delta x} x^{2}$ = $\frac{\delta}{\delta x}[ x + ... + x]$ (x times)
$2x = 1 + ... + 1$ (x times) $= x$ !?

Surely, there must be something wrong... and there is! But I thought this was pretty cool.

 

[bp_psy]

$2^{2} = 2 + 2$
$3^{2} = 3 + 3 + 3$
$4^{2} = 4 + 4 + 4 + 4$

This is true if x is a natural number only.

Generalizing...
$x^{2} = x + ... + x$ (x times)

Take the derivative of both sides.
$\frac{\delta}{\delta x} x^{2}$ = $\frac{\delta}{\delta x}[ x + ... + x]$ (x times)
$2x = 1 + ... + 1$ (x times) $= x$ !?

Surely, there must be something wrong... and there is! But I thought this was pretty cool.

In order to do derivative you need a real variable x and a function for which differentiation makes sense.You should really try to understand what are the conditions for differentiation to avoid this kind of nonsense.

 

[I like Serena]

The derivative of your sequence fails to take into account that an extra term pops up.

Look at it like this:
$$\frac{d}{dx}f(x) \approx {f(x+1)-f(x) \over 1}$$
$$\frac{d}{dx}[x + ... + x] \approx {[(x+1) + ... + (x+1) + (x+1)] - [x + ... + x] \over 1} = 1 + ... + 1 + (x+1) = 2x+1$$Note that it normal to approximate (squeeze) integrals with summations, and summations with integrals.
Some major proofs are built on that concept.

 

[HallsofIvy]

When you write "$x^2= x+ x+ \cdot\cdot\cdot+ x$, x times", the "x times" summation is itself a function of x. You cannot just differentiate the sum as if the number of terms were a constant.